Team Centroid is slacking

Two teams of shovelers plan to remove all the snow from a parking lot that’s shaped like a regular hexagon. Team Vertex initially places each of its six shovelers at the six corners of the lot. Meanwhile, Team Centroid initially places all its shovelers at the very center of the lot.

Each team is responsible for shoveling the snow that is initially closer to someone on their own team than anyone on the other team. What fraction of the lot’s snow is Team Centroid responsible for shoveling?

We can draw a straight line from Team Centroid's central spot to each of the vertices. Drawing a perpendicular bisector of each of these lines we obtain a new smaller, rotated hexagon, shown in blue in the figure below. We note that the apothem of the smaller hexagon is one half of the circumradius of the larger hexagon. Since the area of a regular $n$-gon is given by $$A_n = na^2 \tan \frac{\pi}{n} = \frac{1}{2} nR^2 \sin \frac{2\pi}{n},$$ we see that the apothem and circumradius are related by $a = R \cos \frac{\pi}{n},$ so in this case we have $a = R \cos \frac{\pi}{6} = \frac{R \sqrt{3}}{2},$ or equivalently $R = \frac{2a}{\sqrt{3}}.$

Since we have that the apothem of the Team Centroid hexagon is one half of the circumradius of the Team Vertex hexagon, we have $$R_C = \frac{2}{\sqrt{3}} a_C = \frac{1}{\sqrt{3}} R_V.$$ Since areas scale with the square of the circumradius, we have that the area shoveled by Team Centroid is $\left( \frac{1}{\sqrt{3}} \right)^2 = \frac{1}{3}$ of the entire hexagonal parking lot.